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\title{\bf Additional Vector Analysis}
\author{Airocéan\thanks{airocean@mail.ustc.edu.cn, a@airocean.cn, airocean@foxmail.com, http://airocean.cn}}

\begin{document}

\begin{multicols}{2}
\maketitle
{\bf Proof 1 of dyatic vector.}
\begin{equation}
    \begin{split}
        \vec a \times (\vec b\cdot \vec c)\vec d
        =& \vec a \times \vec d (\vec b \cdot \vec c)\\
        =&\vec e(\vec b \cdot \vec c)\\
        =&(\vec e_2-\vec e_1)[(\vec b_2-\vec b_1)\cdot \vec c]\\
        =&[\vec e_2 \vec b_2-\vec e_2 \vec b_1-\vec e_1 \vec b_2+\vec e_1 \vec b_1]\cdot \vec c.
    \end{split}
\end{equation}

{\bf Proof of $\vec k =-(\vec b \cdot \nabla )\vec b$.}
The basic:
\begin{equation}
    \begin{split}
        \vec b &= \vec \imath b_x+\vec \jmath b_y +\vec k b_z,\\
        \nabla &= \vec \imath \cfrac \partial{\partial x}+\vec \jmath \cfrac \partial{\partial y}+\vec k \cfrac \partial{\partial z}.
    \end{split}
\end{equation}
For $\vec b \cdot \nabla $ as a united operator:
\begin{equation}
    \begin{split}
        \vec b \cdot \nabla= b_x\dfrac\partial{\partial x}+b_y\dfrac\partial{\partial y}+b_z\dfrac\partial{\partial z}.
    \end{split}
\end{equation}The curvature $\vec k$ (not the same $\vec k$ as the unit vector) can be defined as $\mathrm d \vec b / \mathrm d l$ (which is very straightforward to understand):
\begin{equation}
    \begin{aligned}
        \cfrac{\mathrm d \vec b}{\mathrm dl}=\cfrac{\vec \imath \mathrm db_x+\vec \jmath \mathrm db_y+\vec k \mathrm db_z}{\mathrm dl}.
    \end{aligned}
\end{equation}
Multiple the operator $\vec b \nabla$ and $\vec b$, focus on the first term (since the second and the third term use the same approach):
\begin{equation}\label{dier}
    \begin{aligned}
        \vec \imath\left(b_x\dfrac{\partial b_x}{\partial x}+b_y\dfrac{\partial b_x}{\partial y}+b_z\dfrac{\partial b_x}{\partial z}\right).
    \end{aligned}
\end{equation}
We have:
\begin{equation}\label{youhaoei}
    \begin{aligned}
        \mathrm d b_x&=\dfrac{\partial b_x}{\partial x}\mathrm d x+\dfrac{\partial b_x}{\partial y}\mathrm d y+\dfrac{\partial b_x}{\partial z}\mathrm d z,\\
        \dfrac{\mathrm d b_x}{\mathrm d l}&=\dfrac{\partial b_x}{\partial x}\dfrac{\mathrm d x}{\mathrm d l}+\dfrac{\partial b_x}{\partial y}\dfrac{\mathrm d y}{\mathrm d l}+\dfrac{\partial b_x}{\partial z}\dfrac{\mathrm d z}{\mathrm d l}.
    \end{aligned}
\end{equation}
Using the definition of the unit vector $\vec b$ and a very easy triangle relation, we have:
\begin{equation}\label{haoei}
    \begin{aligned}
        \cfrac{\mathrm dx}{\mathrm dl}=b_x=\cfrac{b_x}{|b|},
    \end{aligned}
\end{equation}combine equation \ref{youhaoei} and equation \ref{haoei}, it is very easy to get:
\begin{equation}
    b_x\dfrac{\partial b_x}{\partial x}+b_y\dfrac{\partial b_x}{\partial y}+b_z\dfrac{\partial b_x}{\partial z},
\end{equation}which happens to be the norm of the curvature vector $\vec k$ in the $\vec \imath$ direction, which is shown in equation \ref{dier}. This is also why the equation should be written at the form of $\vec k =-(\vec b \cdot \nabla )\vec b$, instead of \sout{$\vec k =-(\vec b \nabla )\cdot \vec b$} or something else. This will soon be proven next.

{\bf Proof of why the Taylor Formula of a vector should be written at the form of $(\vec r\cdot \nabla)\vec B_0$ instead of $\vec r(\nabla \cdot \vec B_0)$.}

The ordinary Taylor Formula is written as:
\begin{equation}
f(x)=\frac{f\left(x_0\right)}{0 !}+\frac{f^{\prime}\left(x_0\right)}{1 !}\left(x-x_0\right)+\ldots+\frac{f^{(n)}\left(x_0\right)}{n !}\left(x-x_0\right)^n+\ldots
\end{equation}For a multi-variables function:
\begin{equation}
    \begin{aligned}
        f(x,y)=f(x_0,y_0)+&\dfrac{(\partial f/\partial x)^{(1)}}{1!}(x-x_0)+\ldots\\
        +&\dfrac{(\partial f/\partial y)^{(1)}}{1!}(y-y_0)+\ldots
    \end{aligned}
\end{equation}And,
\begin{equation}
    \begin{aligned}
        \vec B[B_x(x,y,z),B_y(x,y,z),B_z(x,y,z)]
        =\vec \imath B_x+\vec \jmath B_y+\vec k B_z
    \end{aligned}
\end{equation}Using Taylor Formula,
\begin{equation}\label{important}
    \begin{aligned}
        B_x(x,y,z)=B_x(x_0,y_0,z_0)&+\cfrac{\partial B_x}{\partial x}(x-x_0)+\ldots\\
        &+\cfrac{\partial B_x}{\partial y}(y-y_0)+\ldots\\
        &+\cfrac{\partial B_x}{\partial z}(z-z_0)+\ldots\\
    \end{aligned}
\end{equation}And for $\vec r(\nabla \cdot \vec B)$,
\begin{equation}
    \begin{aligned}
        \vec r(\nabla \cdot \vec B)=(\vec \imath x+\vec \jmath y+\vec k z)\cdot\left(\dfrac{\partial B_x}{\partial x}+\dfrac{\partial B_y}{\partial y}+\dfrac{\partial B_z}{\partial z}\right),
    \end{aligned}
\end{equation}
and,
\begin{equation}
    \begin{aligned}
        (\vec r\cdot \nabla ) \vec B = \left(x\dfrac{\partial}{\partial x}+y\dfrac{\partial}{\partial y}+z\dfrac{\partial}{\partial z}\right)\cdot(\vec \imath B_x+\vec \jmath B_y+\vec k B_z).
    \end{aligned}
\end{equation}Which one meets the demand of equation \ref{important} is extremely obvious. The proof can always be used in the first order approximation. Like when we try to calculate the magnetic drift in plasma physics.

{\bf Proof of $\nabla \times (\nabla \times \vec A)=\nabla (\nabla \cdot \vec A)-(\nabla \cdot \nabla )\vec A$}. First,
\begin{equation}
    \begin{aligned}
    % \begin{align}
        \nabla \times \vec B&= \left|\begin{array}{ccc}
            \vec{\imath} & \vec{\jmath} & \vec{k} \\
            \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
            B x & B y & B z
            \end{array}\right|\\
             &=\left(\dfrac {\partial B_z}{\partial y}-\dfrac {\partial B_y}{\partial z}\right)\vec \imath+\left(\dfrac {\partial B_x}{\partial z}-\dfrac {\partial B_z}{\partial x}\right)\vec \jmath+\left(\dfrac {\partial B_y}{\partial x}-\dfrac {\partial B_x}{\partial y}\right)\vec k
    % \end{align}
    \end{aligned}
\end{equation}
\begin{equation}
    \nabla \times (\nabla \times \vec A)=\left|\begin{array}{ccc}
            \vec{\imath} & \vec{\jmath} & \vec{k} \\
            \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
            \dfrac{\partial A_z}{\partial y}-\dfrac{\partial A_y}{\partial z} & \dfrac{\partial A_x}{\partial z}-\dfrac{\partial A_z}{\partial x} & \dfrac{\partial A_y}{\partial x}-\dfrac{\partial A_x}{\partial y}
            \end{array}\right|.
\end{equation}
\begin{equation}
    \nabla (\nabla \cdot \vec A)=\left(\vec \imath \dfrac{\partial}{\partial x}+\vec \jmath \dfrac{\partial}{\partial y}+\vec k \dfrac{\partial}{\partial z}\right)\left(\dfrac{\partial A_x}{\partial x}+\dfrac{\partial A_y}{\partial y}+\dfrac{\partial A_z}{\partial z}\right),
\end{equation}
\begin{equation}
    (\nabla \cdot \nabla )\vec A = \left(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}\right)(\vec \imath A_x+\vec \jmath A_y+\vec k A_z).
\end{equation}Then, it is quite easy to use preliminary school calculating method to see their mathematical relation. {\bf All proofs above can be better written in a tensor form.}
% \begin{equation}
% \nabla \phi=\sum_i \dfrac{\partial \phi}{\partial x_i} \hat{e_i}=\partial_i \phi
% \end{equation}
\begin{equation}
\nabla \cdot \boldsymbol{A}=\sum_i \dfrac{\partial A_i}{\partial x_i}=\partial_i A^i.
\end{equation}\begin{equation}
\begin{aligned}
\nabla \times \boldsymbol{A} &=\sum_i\left(\dfrac{\partial A_k}{\partial x_j}-\dfrac{\partial A_j}{\partial x_k}\right) \hat{e_i} \\
&=\sum_{i j k} \varepsilon_{i j k} \dfrac{\partial A_j}{\partial x_i} \hat{e_k} \\
&=\varepsilon^{i j k} \partial_i A_j=\varepsilon^{k i j} \partial_i A_j.
\end{aligned}
\end{equation}

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